Question

When a potential difference of 2 volt is applied across the ends of a wire of 5 metre length a current of 1 ampere flows through it,calculate :-

(i) the resistance per unit length of the wire
(ii) the resistance of 2 metre length of the wire
(iii) the resistance across the ends of the wire if it is doubled on itself

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Answer 1

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(i) 0.4 Ω m^-1

(ii) 0.8 Ω

(iii) 0.5 Ω

Step-by-step explanation:

(i) Given, V = 2 volt, I = 1 A

Resistance of 5 m length of wire :-

`= > R = (V)/(I) = (2)/(1) = 2 \: Ω`

Resistance per unit length of the wire :-

`= > (R)/(l) = (2)/(5) = 0.4 \: Ω \: {m}^( - 1)`

(ii) Resistance of 2m length of the wire~

`= \: 0.4 \: Ω \: {m}^( - 1) * 2 \: m = 0.8 \: Ω`

(iii) When the wire is doubled on itself, its area of cross section becomes twice and the length becomes half. Let a be the initial cross section and ρ be the specific resistance of the material of wire. Then,

`from \ relation \: \: ➪ \: R = ρ * (1)/(a). \\ \\ we \: have, \\ \\ initial \: resistance \:➪ \: 2 = ρ* (5)/(a)....eqn(i) \\ \\ new \: resistance \: ➪ \: R` = ρ* (2.5)/(2a)....eqn(ii) \\ \\ on \: dividing \: eqn(ii) \: by \: eqn(i)... \\ \\ (R`)/(2) = (2.5ρ)/(2a) / (5ρ)/(a) = (1)/(4) \\ \\ R` = (2)/(4) = 0.5 \: Ω`

Thus, on doubling the wire on itself, its resistance becomes one-fourth.

Hope it helps you!!